I am trying to pass a modelclass store to a RenderPartial. The goal for the renderpartial is to change/set values on this (store)model. I have been trying like this:
#{ Html.RenderPartial("test", new store(){Output=""}); }
#{ Html.RenderPartial("test", new store(){Output2=""}); }
public class store
{
public string Output { get; set; }
public string Output2 { get; set; }
}
the Partial 'test' has to change the Output properties. Is it uberhaupt possible and if yes how to do this? The renderpartial contains a javascript to calculate the value of the properties.
RenderPartial is meant to get either none of the data from the parent or a part of the model.
The overloads at:
http://msdn.microsoft.com/en-us/library/system.web.mvc.html.renderpartialextensions.renderpartial(v=vs.108).aspx
Tell you object isn't a custom object but meant to be part of the Model, ex Model.Customers
Pass the required value from your model to the partial and let the partial create it's own objects.
If you really want to pass it to the partial then create a new view model for your parent view and set the Output property in your viewmodel and pass it to the partial.
Note also that the partial gets it's own copy of the data and cannot update the parents copy so that may defeat what you actually want here.
If you need some other calculate data do it in your controller prior to passing it to the view if possible.
#store st = new store(){Output="", Output2=""};
#{ Html.RenderPartial("test", new RouteValueDictionary {{"output", st.Output}, {"output2", st.Output2}}); }
Related
How do I pass a whole model via html.actionlink or using any other method except form submission? Is there any way or tips for it?
Though it's not advisable in complex cases, you can still do that!
public class QueryViewModel
{
public string Search { get; set; }
public string Category { get; set; }
public int Page { get; set; }
}
// just for testing
#{
var queryViewModel = new QueryViewModel
{
Search = "routing",
Category = "mvc",
Page = 23
};
}
#Html.ActionLink("Looking for something", "SearchAction", "SearchController"
queryViewModel, null);
This will generate an action link with href like this,
/SearchController/SearchAction?Search=routing&Category=mvc&Page=23
Here will be your action,
public ViewResult SearchAction(QueryViewModel query)
{
...
}
No, you cannot pass entire complex objects with links or forms. You have a couple of possible approaches that you could take:
Include each individual property of the object as query string parameters (or input fields if you are using a form) so that the default model binder is able to reconstruct the object back in the controller action
Pass only an id as query string parameter (or input field if you are using a form) and have the controller action use this id to retrieve the actual object from some data store
Use session
You could use javascript to detect a click on the link, serialize the form (or whatever data you want to pass) and append it to your request parameters. This should achieve what you're looking to achieve...
I have an action that creates a List and returns it to my view..
public ActionResult GetCustomers()
{
return PartialView("~/Views/Shared/DisplayTemplates/Customers.cshtml", UserQueries.GetCustomers(SiteInfo.Current.Id));
}
And in the "~/Views/Shared/DisplayTemplates/Customers.cshtml" view I have the following:
#model IEnumerable<FishEye.Models.CustomerModel>
#Html.DisplayForModel("Customer")
Then I have in the "~/Views/Shared/DisplayTemplates/Customer.cshtml" view:
#model FishEye.Models.CustomerModel
#Model.Profile.FirstName
I am getting the error:
The model item passed into the dictionary is of type System.Collections.Generic.List`1[Models.CustomerModel]', but this dictionary requires a model item of type 'Models.CustomerModel'.
Shouldn't it display the Customer.cshtml for every item in the collection in the Customers.cshtml?
Help!
I am not sure why you are calling a partial view like this. If it is a Customer Specific view, why not put it under Views/Customer folder ? Remember ASP.NET MVC is more of Conventions. so i would always stick with the conventions (unless abosultely necessary to configure myself) to keep it simple.
To handle this situation, i would do it in this way,
a Customer and CustomerList model/Videmodel
public class CustomerList
{
public List<Customer> Customers { get; set; }
//Other Properties as you wish also
}
public class Customer
{
public string Name { get; set; }
}
And in the action method, i would return an object of CustomerList class
CustomerList customerList = new CustomerList();
customerList.Customers = new List<Customer>();
customerList.Customers.Add(new Customer { Name = "Malibu" });
// you may replace the above manual adding with a db call.
return View("CustomerList", customerList);
Now there should be a view called CustomerList.cshtml under Views/YourControllerName/ folder. That view should look like this
#model CustomerList
<p>List of Customers</p>
#Html.DisplayFor(x=>x.Customers)
Have a view called Customer.cshtml under Views/Shared/DisplayTemplates with this content
#model Customer
<h2>#Model.Name</h2>
This will give you the desired output.
Your view is expecting a single model:
#model FishEye.Models.CustomerModel // <--- Just one of me
You're passing it an anonymous List:
... , UserQueries.GetCustomers(SiteInfo.Current.Id) // <--- Many of me
You should change your view to accept the List or determine which item in the list is supposed to be used before passing it into the View. Keep in mind, a list with 1 item is still a list and the View is not allowed to guess.
Any input much appreciated :)
I want to know one thing whether I can post multiple partial views data in MVC?(means i want to update partial views data to DATABASE)
Here is the Example:
Model:-
public class PassengerViewModel
{
public List<PassengerModel> Passengers { get; set; }
public ContactModel Contact { get; set; }
}
Controller:-
[RequiredAuthentication]
public ActionResult Passenger()
{
var passengrViewMdl = new PassengerViewModel()
{
Contact = new ContactModel(),
Passengers = psngrService.LoadPassengers(Convert.ToInt32(Session["LPORefNO"]))
};
return View(passengrViewMdl);
}
[HttpPost]
public ActionResult Passenger(PassengerViewModel passengerViewModel)
{
Here i want to update Passengers & Contact information
}
View:-
#model QR.LPO.Core.Models.PassengerViewModel
#{
ViewBag.Title = "Add Passengers";
}
#using (Html.BeginForm())
{
#Html.Partial("_Passenger", Model.Passengers);
#Html.Partial("_PassengerContact", Model.Contact);
<input type="submit" value="Submit" />
}
Thanks.
Yes, indeed you can, but, controller usually works only with one model per request, so either your model should have declared within it properties of both partial submodels, or submodels themselves.
This is possible due to HTML specifications, all data on form, which has submit buttom is send to submit action url.
This will almost work as you have it - there's nothing inherent to partials that would prevent this, in the end the html that's output is all that's important.
The problem with your code is that presumably the model of your _Passenger view is of type Passengers and the model of your _PassangerContact view is of type Contact. What this means is that if you standard HtmlHelper extensions (like Html.Textbox(...) or Html.TextboxFor(...) the fields they generate will not have full names like Contact.Name, but instead just names relative to their model, like Name. This will cause modelbinding to fail in your post action.
You can solve this in a number of ways.
Simply use the same model type (PassengerViewModel) in your sub-views, and write code like #Html.TextboxFor(m => m.Contact.Name).
Instead of using Html.Partial, use Html.EditorFor(...). This passes the proper prefix information into the child view so the field names are generated properly.
Explicitly set the prefix yourself
Like this:
#{
var childViewData = new ViewDataDictionary(this.ViewData);
childView.TemplateInfo.HtmlFieldPrefix = "Contact";
}
#Html.Partial("_PassengerContact", Model.Contact, childViewData)
You could also look at creating a Html.PartialFor overload yourself, as described in this stackoverflow question: ASP.NET MVC partial views: input name prefixes
Getting my head around MVC today and ran across the best practice of not passing a Model directly to a view. Instead, use a ViewModel.
I researched AutoMapper and plan on using it to map my ViewModels to the respective Models. And I understand that AutoMapper is smart enough to map IEnumerable to IEnumerable without a separate mapping, as long as source and dest are mapped.
But I'm a bit confused about how to handle passing an IEnumerable in my ViewModel to my view. I currently have my page working using a ViewModel that includes IEnumerable but I read that this is just as bad as passing the IEnumerable directly to the view. So do I need a separate ViewModel to hold the object which will be used in an IEnumerable property of the main ViewModel?
So where Activity is the Model in question:
public class ActivityHistoryViewModel
{
public IEnumerable<Activity> activities { get; set; }
}
do I need to create ActivityViewModel and write my ActivityHistoryViewModel like this?
public class ActivityHistoryViewModel
{
public IEnumerable<ActivityViewModel> activities { get; set; }
}
Is there an easier way to do this?
Yes, this is correct. Assuming the only data your model will need is the list, then you don't really need ActivityHistoryViewModel and the view can be typed as such:
#model IEnumerable<ActivityViewModel>
your auto mapper config would look like this:
Mapper.CreateMap<Activity, ActivityViewModel>();
you would map like this:
IEnumerable<Activity> data = GetActivities();
var model = Mapper.Map<IEnumerable<Activity>, IEnumerable<ActivityViewModel>>(data);
return View(model);
And when you define ActivityViewModel you can either create a property-for-property duplicate type, or trim out the excess data you don't need (in my case it would be something like "created date", that is db generated and of no importance to users).
Or if you want to stick with ActivityHistoryViewModel to pass along more than just the list:
view type:
#model ActivityHistoryViewModel
mapping config can remain the same
map like this:
IEnumerable<Activity> data = GetActivities();
var model = new ActivityHistoryViewModel() {
someOtherProperty = "hello world!",
activities = Mapper.Map<IEnumerable<Activity>, IEnumerable<ActivityViewModel>>(data)
};
return View(model);
Questions
There are actually two related questions:
Should I create a ViewModel for each page?
If you do not have problems in creating a single ViewModel class for the two pages (Create.cshtml and Edit.cshtml) how can I validate the ViewModel in different ways (depending on the page that is being used)
Source
ViewModel
public class ProjectViewModel
{
public string Name { get; set; }
public string Url { get; set; }
public string Description { get; set; }
}
Edit.cshtml
#using BindSolution.ViewModel.Project
#model ProjectViewModel
#{
ViewBag.Title = Model.Name;
}
#Html.EditorForModel()
Create.cshtml
#using BindSolution.ViewModel.Project
#model ProjectViewModel
#{
ViewBag.Title = "New Project";
}
#Html.EditorForModel()
ProjectValidator.cs
public class ProjectValidator : AbstractValidator<ProjectViewModel>
{
private readonly IProjectService _projectService;
public ProjectValidator(IProjectService projectService)
{
_projectService = projectService;
RuleFor(p => p.Name)
.NotEmpty().WithMessage("required field")
/*The validation should be made only if the page is Create.cshtml. That is, if you are creating a new project.*/
.When(p => p.??) //Problem Here!!
.Must(n => !_projectService.Exist(n)).WithMessage("name already exists");
RuleFor(p => p.Url)
.NotEmpty().WithMessage("required field");
}
}
Note that if the user is editing an existing project, validation of the property name should not be done again.
ProjectController.cs > Edit method
[HttpPost]
public ActionResult Edit(Guid projectID, ProjectViewModel model)
{
var project = _projectService.Repository.Get(projectID);
if (ModelState.IsValid && TryUpdateModel(project))
{
_projectService.Repository.Attach(project);
if (_projectImageWrap.Create(project) && _projectService.Repository.Save() > 0)
return AjaxRedirect("Index");
}
return View(model);
}
Notes
If I create a ViewModel for each page, there is a duplication of code since pages have the same properties.
Add a property on the ViewModel indicating what page it is being displayed does not solve my problem as to instantiate the ViewModel, I use AutoMapper.
To validate the data, I use FluentValidator.
Thank you all for your help!
My understanding is that there isn't a 1:1 correlation between ViewModels and Views. Oftentimes you will have a View that will not require a ViewModel to go alongside with it.
You will want to create a ViewModel if and only if you need a Model absolutely paralleled and tailored to a specific View. This will not be the case 100% of the time.
When the functionality / use case /validation is different between the pages I use different models. If its the exact same besides the presence of an ID or something similar I use the same model, and its also possible to just use the same view if the differences are pretty minor.
Since your validation is different, if I were doing it I would create two different models so that I could use the out of the box DataAnnotations, with your validation though it may not be required. You could also on the edit model have a readonly property for name since its not editable any longer.
For me the same object must have the same validation on every time, in main to ensure the consistence of the object, independently if it was created or edited.
i think that you should create only one validation, and edit your "exists" method to pass to verify if it is a new object or the current object in repository.
Personally, I don't have a problem with 2 view models, especially if (as Paul Tyng suggested) you use a base class for the fields that are common to edit and create scenarios.
However, if you really only want a single view model then you would either need to:
add a flag to the view model and use the When() method in your validator. Note though that this will not generate the appropriate client-side only validation
define a second validator and invoke the appropriate one from the controller (i.e. instead of the "automatic" validation)
Provide another view Edit.cshtml which will allow the user to edit the data for a selected item.
Create another view Query.cshtml which based on the ItemName will allow the users to query the Inventory table.
Perform the calculation for the total profit (numbersold times (saleprice-purchasecost). Display the total profit.
(BONUS) Create another view Sell.cshtml that will indicate the sale of an item. Adding one to NumberSold and subtract one from NumberInventory for the selected record.